[LeetCode] Number of Arithmetic Triplets

1.문제

You are given a 0-indexed, strictly increasing integer array nums and a positive integer diff. A triplet (i, j, k) is an arithmetic triplet if the following conditions are met:

• i < j < k,
• nums[j] - nums[i] == diff, and
• nums[k] - nums[j] == diff.

Return the number of unique arithmetic triplets.

index가 0부터 시작하는 숫자 배열이 주어질 때 다음 조건을 만족하는 triplet의 갯수를 리턴하는 문제이다.

• i < j < k,
• nums[j] - nums[i] == diff, and
• nums[k] - nums[j] == diff.

Example 1

Input: nums = [0,1,4,6,7,10], diff = 3
Output: 2
Explanation:
(1, 2, 4) is an arithmetic triplet because both 7 - 4 == 3 and 4 - 1 == 3.
(2, 4, 5) is an arithmetic triplet because both 10 - 7 == 3 and 7 - 4 == 3. 

Example 2

Input: nums = [4,5,6,7,8,9], diff = 2
Output: 2
Explanation:
(0, 2, 4) is an arithmetic triplet because both 8 - 6 == 2 and 6 - 4 == 2.
(1, 3, 5) is an arithmetic triplet because both 9 - 7 == 2 and 7 - 5 == 2.

Constraints:

• 3 <= nums.length <= 200
• 0 <= nums[i] <= 200
• 1 <= diff <= 50
• nums is strictly increasing.

2.풀이

1. 3중 for문을 돌면서 조건을 체크한다.

/**
 * @param {number[]} nums
 * @param {number} diff
 * @return {number}
 */
const arithmeticTriplets = function (nums, diff) {
  let count = 0;
  for (let i = 0; i < nums.length - 2; i++) {
    for (let j = i + 1; j < nums.length - 1; j++) {
      for (let k = j + 1; k < nums.length; k++) {
        // for문을 돌면서 조건에 맞는지 체크한다.
        if (nums[j] - nums[i] === diff && nums[k] - nums[j] === diff) {
          count++;
        }
      }
    }
  }

  return count;
};

3.결과